package com.freetymekiyan.algorithms.level.hard;

/**
 * 321. Create Maximum Number
 * <p>
 * Given two arrays of length m and n with digits 0-9 representing two numbers. Create the maximum number of length k <=
 * m + n from digits of the two. The relative order of the digits from the same array must be preserved. Return an array
 * of the k digits.
 * <p>
 * Note: You should try to optimize your time and space complexity.
 * <p>
 * Example 1:
 * <p>
 * Input:
 * nums1 = [3, 4, 6, 5]
 * nums2 = [9, 1, 2, 5, 8, 3]
 * k = 5
 * Output:
 * [9, 8, 6, 5, 3]
 * Example 2:
 * <p>
 * Input:
 * nums1 = [6, 7]
 * nums2 = [6, 0, 4]
 * k = 5
 * Output:
 * [6, 7, 6, 0, 4]
 * Example 3:
 * <p>
 * Input:
 * nums1 = [3, 9]
 * nums2 = [8, 9]
 * k = 3
 * Output:
 * [9, 8, 9]
 * <p>
 * Companies: Google
 * <p>
 * Related Topics: Dynamic Programming, Greedy
 * <p>
 * Similar Questions: (M) Remove K Digits, (M) Maximum Swap
 */
public class CreateMaximumNumber {

  public int[] maxNumber(int[] nums1, int[] nums2, int k) {
    int n = nums1.length;
    int m = nums2.length;
    int[] ans = new int[k];
    for (int i = Math.max(0, k - m); i <= k && i <= n; ++i) {
      int[] candidate = merge(maxArray(nums1, i), maxArray(nums2, k - i), k);
      if (greater(candidate, 0, ans, 0)) ans = candidate;
    }
    return ans;
  }

  private int[] merge(int[] nums1, int[] nums2, int k) {
    int[] ans = new int[k];
    for (int i = 0, j = 0, r = 0; r < k; ++r)
      ans[r] = greater(nums1, i, nums2, j) ? nums1[i++] : nums2[j++];
    return ans;
  }

  public boolean greater(int[] nums1, int i, int[] nums2, int j) {
    while (i < nums1.length && j < nums2.length && nums1[i] == nums2[j]) {
      i++;
      j++;
    }
    return j == nums2.length || (i < nums1.length && nums1[i] > nums2[j]);
  }

  public int[] maxArray(int[] nums, int k) {
    int n = nums.length;
    int[] ans = new int[k];
    for (int i = 0, j = 0; i < n; ++i) {
      while (n - i + j > k && j > 0 && ans[j - 1] < nums[i]) j--;
      if (j < k) ans[j++] = nums[i];
    }
    return ans;
  }
}